如何通过其运行时类型抛出异常？

Question

I want to call a function that may throw an exception. If it does throw an exception, I want to catch it and pass the exception object to a handler function. The default implementation of the handler function is simply to throw the exception. Here is whittled-down code to illustrate the issue:

struct base_exception : exception {
  char const* what() const throw() { return "base_exception"; }
};

struct derived_exception : base_exception {
  char const* what() const throw() { return "derived_exception"; }
};

void exception_handler( base_exception const &e ) {
  throw e; // always throws a base_exception object even if e is a derived_exception
}

int main() {
  try {
    throw derived_exception();
  }
  catch ( base_exception const &e ) {
    try {
      cout << e.what() << endl; // prints "derived_exception" as expected
      exception_handler( e );
    }
    catch ( base_exception const &e ) {
      cout << e.what() << endl; // prints "base_exception" due to object slicing
    }
  }
}

However, the throw e in exception_handler() throws a copy of the static type of the exception, ie, base_exception . How can I make exception_handler() throw the actual exception having the correct run-time type of derived_exception ? Or how can I redesign things to get what I want?

Answer 1

You can put a throw_me virtual function in the base exception class, and have every derived class override it. The derived classes can throw the proper most derived type, without slicing. Even though the function has the same definition in each class, they're not the same - the type of *this is different in each case.

struct base_exception : exception
{
  char const* what() const throw() { return "base_exception"; }
  virtual void throw_me() const { throw *this; }
};

struct derived_exception : base_exception
{
  char const* what() const throw() { return "derived_exception"; }
  virtual void throw_me() const { throw *this; }
};

void exception_handler( base_exception const &e ) {
  e.throw_me();
} 

Answer 2

You can use throw; to re-throw the exception that was caught. You could also use a template.

template<typename T> void rethrow(const T& t) { throw t; }

Answer 3

Throw by value, catch by reference . It'll save you a lot of headaches.

Answer 4

What you are looking for is called "propagating" the exception. To do so, you have to use the throw keyword without parameters inside the catch block. It will not copy the exception and the exception will be caught by the next catch block on its way or will make your program abort if it's not caught again.